Originally Posted by

**mathaddict** hi punch ,

for (1) , you multiply the whole thing by x so you don end up with $\displaystyle x^{-1}$

$\displaystyle mx-3=x+\frac{1}{x}$

$\displaystyle mx^2-3x=x^2+1$

$\displaystyle (m-1)x^2-3x-1=0$ **

If a line is a tangent to the curve , it touches the curve at one point .

so $\displaystyle b^2-4ac=0$

$\displaystyle (-3)^2-4(m-1)(-1)=0$

calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .

for (2) , you are given the range , so

$\displaystyle (x+5)(x-3)<0\Rightarrow x^2+2x-15<0$

compared this with , $\displaystyle x^2+cx-d<0$ ?? (i think you have a typo here)

we see that c=2 , d=15 assuming that i guess it correctly .